# Electromagnetic induction of an uniform magnetic field

EDIT: Thanks to the first answer, I may see that there is differences between a giant solenoid and a completely uniform magnetic field. Therefore it would be great if one can explain to me both the case.

Let's say there is an uniform magnetic field inside a very long and large solenoid. The magnetic field is slowly decreased. How can we find the electric field?

From the Maxwell-Faraday Equation we have $\nabla \times E=-\frac{\partial B}{\partial t}$ and it's can be solved by a curly electric field with the value of electric field is constant (proportional to $|\frac{\partial B}{\partial t}|$) and directed to circles around a centre. But the centre is arbitrarily (one may say it's the central of the solenoid but I find it uncompelling since any centre will satisfy Maxwell equation). So how can one find the electric field in this situation?

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"can be solved by a curly electric field with the value of electric field is constant (proportional to $|\frac{\partial B}{\partial t}|$) and directed to circles around a centre." That's not true. The electric field has magnitude $E(r) = \frac{k}{2}r$, proportional to distance $r$ from the center axis. –  Ján Lalinský Aug 16 at 7:07
@JánLalinský Your solution is correct inside the solenoid. But there is also an outside, lol... –  MariusMatutiae Aug 16 at 8:13

Uniformity of $\mathbf B$ in space and the equation $$\nabla \times \mathbf E = -\frac{\partial\mathbf B}{\partial t}$$ implies uniformity of $\nabla \times \mathbf E$, but not uniformity of $\mathbf E$, because these conditions do not determine electric field uniquely - there is infinity of solutions. It is an artificial situation which lacks further conditions and no unique electric field can selected.

In case we are interested in the fields of an isolated cylindrical solenoid, there is a new condition: the electric field has the cylindrical symmetry determined by the solenoid. Together with some other conditions, like continuity of the electric field and limiting magnitude 0 at infinity, unique electric field may be selected, which is centered on the solenoid.

Uniqueness of this solution above all other solutions of the above differential equation can be easily understood in terms of retarded fields; if we assume that the EM fields are determined by the charge and current density, since these can be non-zero only in the winding of the solenoid, the EM field has to have cylindrical symmetry of the solenoid.

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There is no need to impose boundary for $\mathbf E$. It is derived to be $\propto\frac{1}{r}$ outside of the infinitely long solenoid from the fact the magnetic field vanishes outside the solenoid. –  Hans Aug 16 at 13:54
@Hans, I've changed the answer. –  Ján Lalinský Aug 17 at 0:44
That is better. –  Hans Aug 17 at 4:56